Left-uniformly continuous bounded functions (UCB)

If $\phi \in \mathrm{U}\mathrm{C}\mathrm{B}(G)$, then it is continuous at every $t\in G$. Indeed, for $y\to t$ put $x=y{t}^{-1}\to e$; then $\phi (y)=\phi (xt)=(L_{x^{-1}}\phi )(t)$, so $\mathrm{\mid }\phi (y)-\phi (t)\mathrm{\mid }\le \mathrm{\parallel }{L}_{x^{-1}}\phi -\phi {\mathrm{\parallel }}_{\sup }\to 0$. Here $x^{-1}\to e$ by continuity of inversion.

The zero function belongs to $\mathrm{U}\mathrm{C}\mathrm{B}(G)$. For $\phi ,\psi \in \mathrm{U}\mathrm{C}\mathrm{B}(G)$ and $a\in \mathbb{C}$, the estimates $\mathrm{\parallel }{L}_x(\phi +\psi )-(\phi +\psi ){\mathrm{\parallel }}_{\sup }\le \mathrm{\parallel }{L}_x\phi -\phi {\mathrm{\parallel }}_{\sup }+\mathrm{\parallel }{L}_x\psi -\psi {\mathrm{\parallel }}_{\sup }$ and $\mathrm{\parallel }{L}_x(a\phi )-a\phi {\mathrm{\parallel }}_{\sup }=\mathrm{\mid }a\mathrm{\mid }\mathrm{\parallel }{L}_x\phi -\phi {\mathrm{\parallel }}_{\sup }$ show closure under addition and scalar multiplication. Thus it is a linear subspace of the bounded functions.

Let $\phi$ lie in the norm closure of $\mathrm{U}\mathrm{C}\mathrm{B}(G)$. For $x\in G$, choose $\psi \in \mathrm{U}\mathrm{C}\mathrm{B}(G)$ with $2\mathrm{\parallel }\phi -\psi {\mathrm{\parallel }}_{\sup }<\epsilon \mathrm{/}2$. Then $\mathrm{\parallel }{L}_x\phi -\phi {\mathrm{\parallel }}_{\sup }\le 2\mathrm{\parallel }\phi -\psi {\mathrm{\parallel }}_{\sup }+\mathrm{\parallel }{L}_x\psi -\psi {\mathrm{\parallel }}_{\sup }$, since left translation is an isometry for the sup norm. By the defining condition for $\psi$, a neighborhood of $e$ makes the last term $<\epsilon \mathrm{/}2$. Hence $\mathrm{\parallel }{L}_x\phi -\phi {\mathrm{\parallel }}_{\sup }<\epsilon$ there, so $\phi \in \mathrm{U}\mathrm{C}\mathrm{B}(G)$ and the subspace is closed.

For $g\in G$, the group law gives $L_x{L}_g=L_g{L}_{g^{-1}xg}$. Therefore $\mathrm{\parallel }{L}_x(L_g\phi )-L_g\phi {\mathrm{\parallel }}_{\sup }=\mathrm{\parallel }{L}_{g^{-1}xg}\phi -\phi {\mathrm{\parallel }}_{\sup }\to 0$ as $x\to e$, because $g^{-1}xg\to e$. So $L_g\phi \in \mathrm{U}\mathrm{C}\mathrm{B}(G)$ and the subspace is translation-invariant.

Fix $(x_0,{\phi }_0)\in G\times \mathrm{U}\mathrm{C}\mathrm{B}(G)$. For all $x\in G$ and $\phi \in \mathrm{U}\mathrm{C}\mathrm{B}(G)$, $\mathrm{\parallel }{L}_x\phi -L_{x_0}{\phi }_0{\mathrm{\parallel }}_{\sup }\le \mathrm{\parallel }\phi -{\phi }_0{\mathrm{\parallel }}_{\sup }+\mathrm{\parallel }{L}_{x_0^{-1}x}{\phi }_0-{\phi }_0{\mathrm{\parallel }}_{\sup }$. The first term tends to zero as $\phi \to {\phi }_0$ and the second as $x\to x_0$ by the defining condition for ${\phi }_0$. This proves joint continuity of the left action.

By [F2], each $\phi \in \mathrm{U}\mathrm{C}\mathrm{B}(G)$ is Borel measurable and so defines a class $[\phi ]$ in [F1]. Put $M=\mathrm{\parallel }\phi {\mathrm{\parallel }}_{\sup }$. If $0\le t<M$, then some point has $\mathrm{\mid }\phi \mathrm{\mid }>t$, and continuity makes $\{y:\mathrm{\mid }\phi (y)\mathrm{\mid }>t\}$ a nonempty open set. It has positive measure by [A2], so $\mathrm{\parallel }[\phi ]{\mathrm{\parallel }}_{\mathrm{\infty }}\ge t$. Also $\mathrm{\parallel }[\phi ]{\mathrm{\parallel }}_{\mathrm{\infty }}\le M$ because $\mathrm{\mid }\phi \mathrm{\mid }\le M$ everywhere. Letting $t\uparrow M$ when $M>0$, and noting both norms are zero when $M=0$, gives $\mathrm{\parallel }[\phi ]{\mathrm{\parallel }}_{\mathrm{\infty }}=\mathrm{\parallel }\phi {\mathrm{\parallel }}_{\sup }$. The class map is therefore isometric and injective; in particular, an almost-everywhere zero UCB function is identically zero.